Question: Solve for $x$ and $y$ using substitution. ${x-2y = -1}$ ${x = 5y-10}$
Since $x$ has already been solved for, substitute $5y-10$ for $x$ in the first equation. ${(5y-10)}{- 2y = -1}$ Simplify and solve for $y$ $5y-10 - 2y = -1$ $3y-10 = -1$ $3y-10{+10} = -1{+10}$ $3y = 9$ $\dfrac{3y}{{3}} = \dfrac{9}{{3}}$ ${y = 3}$ Now that you know ${y = 3}$ , plug it back into $\thinspace {x = 5y-10}\thinspace$ to find $x$ ${x = 5}{(3)}{ - 10}$ $x = 15 - 10$ ${x = 5}$ You can also plug ${y = 3}$ into $\thinspace {x-2y = -1}\thinspace$ and get the same answer for $x$ : ${x - 2}{(3)}{= -1}$ ${x = 5}$